For the first time, the FMC will have solvers in the "NxNxN" category for up to 9x9x9 competing to solve 10 random scrambles in as few moves as possible. Also of note: There are now four solvers in the 5x5x5 section.

What are the known bounds for the diameter of this subgroup? This thread gives a lower bound of 26 QTM but doesn't discuss HTM. I seem to recall a lower bound of 20 HTM from somewhere, but no upper bound. However, I recently found a position that requires 21 HTM:

U R F2 R U2 F2 U' F' R U F U R' F R2 U F R U2 R F'

Is this the first one known? Out of all 186624 positions with all pieces correctly permuted, this position and the inverse are the only ones requiring 21. Here's the full distribution:

In the future I think it would be a good idea if we all used URLs of the form:

http://forum.cubeman.org/?q=node/view/563#comment

rather than

http://cubezzz.dyndns.org/?q=node/view/563#comment

dyndns.org has raised their prices every year and I'm considering
dropping their service. Unfortunately if we do that a lot of old URLs will stop
working, so I'm open to any clever ideas on what is the best way to deal with
this problem. I'm committed to keeping the maxhost.org and
cubeman.org domain names working for the long term, but
I'm very unhappy with dyndns.

I optimally solved three million positions in four distinct metrics.
These positions are distinct from the three million positions I ran
some years back. Random numbers were generated with the Mersenne
Twister algorithm. The four metrics I ran were quarter-turn metric,
half-turn metric, slice-turn metric, and axial-turn metric (equivalent
to the robot-turn or simultaneous-turn metric on the 3x3 cube).

The generators for each metric are strict super- or sub-sets of the
generators for the other metrics. The quarter-turn metric has 12
generators, the half-turn metric has 18 generators, the slice-turn

Following on from a highly symmetric presentation I supplied for the miniature Rubiks cube group
Presentation for the 2x2x2 Rubiks cube group
I investigated whether the Mathieu Group M24 could be similarily presented taking full advantage of the
plentiful symmetry inherent in the Rubiks Revenge cube (4x4x4). The answer was indeed yes - here is the presentation I found - again on three involutions:

Write the puzzle as the group <R,F,U,D> where R and F are 180 degree moves. We use a two-phase algorithm to first reduce the state of the puzzle to the subgroup <R3,F3,U,D>, and then finish the solve in the second phase. The subgroup <R3,F3,U,D> is the group of all positions where all of the gears are oriented, because R3 is the same as R' except the gear orientation remains unchanged.

The first phase is easy to compute. There are only 3^8 = 6561 positions because each gear has only 3 different orientations, despite having 6 teeth.

The second phase is harder. The number of positions is 24*8!^2/2 = 19,508,428,800, since it turns out that the permutation of the 3 unfixed edges on the E slice is completely determined by the permutation of the centres. This phase was solved with a BFS and took around 7 and a half hours to complete.

Last night, I found this thread on the speedsolving forums which proves an upper bound of 46 moves. First, the puzzle is separated into two halves, which takes at most 6 moves. Each half is then solved in at most 20 moves (= 7 moves for orientation + 13 moves for permutation, after orientation is solved), for a total of 6+2*(7+13) = 46. xyzzy writes

The ⟨U,R,F⟩ subgroup, while much smaller than G_0, is still pretty large, having 36 billion states. It's small enough that a
full breadth-first search can be done if symmetry+antisymmetry reduction is used, but I will leave this for another time.

Here I'm using sign notation, so 2R is the inner slice only. There are 10 edges, 10 centres in sets of 2, 2, 2 and 4, and 4 permutations of the corner pieces for a total of 4*10!*10!/(2!2!2!4!) = 274,337,280,000 positions. From July 4th 2017 to July 6th 2017, I ran a complete breadth first search of this puzzle in around 60 hours. God's number is 28.